Respuesta :
Answer:
[tex]P(y = 2) = 0.294[/tex]
Step-by-step explanation:
If the times for service calls follow a normal distribution with mean μ = 60 minutes and standar deviation σ = 10 minutes, then we can calculate the probability that one call take more than 68.4 minutes, thus:
P(x>68.4) = 1 - P(X<68.4)
We standardize as follow
[tex]z = \frac{x-\mu}{\sigma}[/tex]
[tex]z = \frac{68.4 - 60}{10} = 0.84[/tex]
Since the table of the distribution normal, we obtain:
1- P(z <0.84) = 1 - 0.7995 = 0.2005
Therefore:
P(x>68.4) = 0.2005
Now if you select random eight service calls, the number of calls takes more than 68.4 minutes is a binomial variable with
p = 0.2005
n = 8
And we need calculate this probability
[tex]P(y = 2) = (8C2)(0.2005)^{2}(1-2005)^{8-2}[/tex]
[tex]P(y = 2) = 0.294[/tex]
